c++ - 重载 operator<< 时 std::endl 的类型未知

Question

我重载了运算符<<

template <Typename T>
UIStream& operator<<(const T);

UIStream my_stream;
my_stream << 10 << " heads";

有效但:

my_stream << endl;

给出编译错误:

error C2678: binary '<<' : no operator found which takes a left-hand operand of type 'UIStream' (or there is no acceptable conversion)

制作 my_stream << endl 的方法是什么？工作吗？

Answer 1

std::endl是一个函数，std::cout通过实现 operator<< 来利用它获取与 std::endl 具有相同签名的函数指针.

在那里，它调用函数，并转发返回值。

这是一个代码示例:

#include <iostream>

struct MyStream
{
    template <typename T>
    MyStream& operator<<(const T& x)
    {
        std::cout << x;

        return *this;
    }


    // function that takes a custom stream, and returns it
    typedef MyStream& (*MyStreamManipulator)(MyStream&);

    // take in a function with the custom signature
    MyStream& operator<<(MyStreamManipulator manip)
    {
        // call the function, and return it's value
        return manip(*this);
    }

    // define the custom endl for this stream.
    // note how it matches the `MyStreamManipulator`
    // function signature
    static MyStream& endl(MyStream& stream)
    {
        // print a new line
        std::cout << std::endl;

        // do other stuff with the stream
        // std::cout, for example, will flush the stream
        stream << "Called MyStream::endl!" << std::endl;

        return stream;
    }

    // this is the type of std::cout
    typedef std::basic_ostream<char, std::char_traits<char> > CoutType;

    // this is the function signature of std::endl
    typedef CoutType& (*StandardEndLine)(CoutType&);

    // define an operator<< to take in std::endl
    MyStream& operator<<(StandardEndLine manip)
    {
        // call the function, but we cannot return it's value
        manip(std::cout);

        return *this;
    }
};

int main(void)
{
    MyStream stream;

    stream << 10 << " faces.";
    stream << MyStream::endl;
    stream << std::endl;

    return 0;
}

希望这能让您更好地了解这些事情的工作原理。